IB Maths AA HL · Unit 3: Geometry and Trigonometry
IB Maths AA HL Vector Lines Questions
Exam-style IB Maths AA HL vector lines questions with worked solutions. Start with the 3 fully worked examples below — each is solved step by step the way an IB examiner expects — then try the practice questions and check your working against the mark scheme.
- 37 questions
- Paper 1: 24
- Paper 2: 13
- 7 easy
- 11 medium
- 15 hard
- 4 starter
- 3 worked examples
Practise Vector Lines questions →
AA HL formula booklet
What's examined in AA HL vector lines
The question bank covers these vector lines question types (number of questions in brackets):
- Vector Product Applications (16)
- Vector Line Equations & Intersection (15)
- Line Distances & Angles (6)
Key formulas
- Vector magnitude
- \(|\vec{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2}\)
- Vector equation of a line
- \(\vec{r} = \vec{a} + t\,\vec{d}\)
- Dot product
- \(\vec{a}\cdot\vec{b} = a_1 b_1 + a_2 b_2 + a_3 b_3 = |\vec{a}||\vec{b}|\cos\theta\)
In the same notation as the IB formula booklet. All AA HL formulas →
Vector Lines worked examples
Worked example 1: The vector product and area · easy
Given the vectors $\mathbf{a} = \mathbf{i} + 2\mathbf{j} - \mathbf{k}$ and $\mathbf{b} = 3\mathbf{i} + \mathbf{k}$, find the exact area of the parallelogram formed by these two vectors.
1. Recognize that the area of a parallelogram defined by two vectors is exactly the magnitude of their vector cross product: Area $= |\mathbf{a} \times \mathbf{b}|$.
2. Write the vectors in column format for easier calculation: $\mathbf{a} = \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix} 3 \\ 0 \\ 1 \end{pmatrix}$.
3. Compute the components of the cross product: $x = (2)(1) - (-1)(0) = 2$.
4. Compute the $y$-component (remembering the negative sign): $y = -((1)(1) - (-1)(3)) = -(1 - (-3)) = -4$.
5. Compute the $z$-component: $z = (1)(0) - (2)(3) = -6$. So, $\mathbf{a} \times \mathbf{b} = \begin{pmatrix} 2 \\ -4 \\ -6 \end{pmatrix}$.
6. Calculate the magnitude to find the area: $\sqrt{2^2 + (-4)^2 + (-6)^2} = \sqrt{4 + 16 + 36} = \sqrt{56} = \mathbf{2\sqrt{14}}$.
Examiner tip: A common mistake is forgetting the negative sign on the middle ($\mathbf{j}$) component when calculating the cross product manually via the determinant method.
Worked example 2: Angle between two vector lines · medium
Find the acute angle between the line $L_1: \mathbf{r} = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} + t \begin{pmatrix} 2 \\ -1 \\ 2 \end{pmatrix}$ and the line $L_2: \mathbf{r} = \begin{pmatrix} 0 \\ -2 \\ 3 \end{pmatrix} + s \begin{pmatrix} 1 \\ 4 \\ -1 \end{pmatrix}$. Give your answer in degrees to one decimal place.
1. Identify the direction vectors of the two lines, ignoring the position vectors: $\mathbf{d}_1 = \begin{pmatrix} 2 \\ -1 \\ 2 \end{pmatrix}$ and $\mathbf{d}_2 = \begin{pmatrix} 1 \\ 4 \\ -1 \end{pmatrix}$.
2. Calculate the scalar product: $\mathbf{d}_1 \cdot \mathbf{d}_2 = (2)(1) + (-1)(4) + (2)(-1) = 2 - 4 - 2 = -4$.
3. Find the magnitude of each direction vector: $|\mathbf{d}_1| = \sqrt{2^2 + (-1)^2 + 2^2} = \sqrt{9} = 3$, and $|\mathbf{d}_2| = \sqrt{1^2 + 4^2 + (-1)^2} = \sqrt{18} = 3\sqrt{2}$.
4. Substitute these values into the angle formula, using the absolute value of the dot product to find the acute angle: $\cos \theta = \frac{|-4|}{3 \times 3\sqrt{2}} = \frac{4}{9\sqrt{2}}$.
5. Evaluate the inverse cosine on your GDC: $\theta = \arccos\left(\frac{4}{9\sqrt{2}}\right) = 71.681...^\circ$.
6. State the final angle: $\mathbf{71.7^\circ}$.
Examiner tip: Only use the direction vectors (the multipliers of $t$ and $s$) to find the angle between lines; the position vectors (starting points) are completely irrelevant to the angle and plugging them into the formula is a frequent conceptual error.
Worked example 3: Proving lines are skew · hard
Line $L_1$ has vector equation $\mathbf{r} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} + t \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix}$ and line $L_2$ has equation $\mathbf{r} = \begin{pmatrix} 2 \\ 1 \\ 0 \end{pmatrix} + s \begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix}$. Show analytically that $L_1$ and $L_2$ are skew lines.
1. Check if the lines are parallel. Since the direction vector $\begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix}$ is not a scalar multiple of $\begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix}$, the lines are not parallel.
2. Equate the $x$ and $y$ parametric components of the lines to test for intersection points: $1 + t = 2 + 2s \implies t - 2s = 1$ (Equation 1), and $2 - t = 1 + s \implies -t - s = -1 \implies t + s = 1$ (Equation 2).
3. Solve this simultaneous system. Subtracting Eq 2 from Eq 1 gives $-3s = 0 \implies s = 0$. Substituting $s=0$ into Eq 2 yields $t = 1$.
4. Substitute these parameter values into the $z$-components of each respective line to see if they match. For $L_1$ ($t=1$): $z = 3 + (1) = 4$.
5. Substitute for $L_2$ ($s=0$): $z = 0 + (-1)(0) = 0$.
6. Conclude: Because the $z$-coordinates do not match ($4 \neq 0$), the lines do not intersect. Since they are neither parallel nor intersecting, they must be skew lines.
Examiner tip: You must explicitly state and verify that the direction vectors are not scalar multiples of each other before testing for intersection; otherwise, you haven't fully proven they are skew rather than just coincident or parallel.
Try these IB Maths AA HL vector lines questions
Three questions from the bank, easiest first. Mark schemes and AI marking of your written working are in the practice area.
Question 1 · easy · 3 marks · Paper 1
Find the vector product \(\mathbf{a} \times \mathbf{b}\) where \(\mathbf{a} = \begin{pmatrix} 1 \\ -2 \\ 3 \end{pmatrix}\) and \(\mathbf{b} = \begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix}\).
Question 2 · medium · 6 marks · Paper 1
Consider the two lines defined by:
\[L_1: \mathbf{r} = \begin{pmatrix} 2 \\ 0 \\ -1 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix} \quad \text{and} \quad L_2: \mathbf{r} = \begin{pmatrix} 3 \\ 2 \\ 0 \end{pmatrix} + \mu \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix}\]
Show algebraically that \(L_1\) and \(L_2\) intersect, and find the coordinates of their point of intersection.
Question 3 · hard · 7 marks · Paper 2
Two lines $L_1$ and $L_2$ are given by the vector equations:
$L_1: \mathbf{r} = \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix} + \lambda \begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix}$
$L_2: \mathbf{r} = \begin{pmatrix} -1 \\ 2 \\ 1 \end{pmatrix} + \mu \begin{pmatrix} 1 \\ -1 \\ 3 \end{pmatrix}$
(a) Show that the lines $L_1$ and $L_2$ are skew.
(b) Find the shortest distance between the lines $L_1$ and $L_2$. Give your answer in the form $\frac{a}{\sqrt{b}}$ where $a, b \in \mathbb{Z}^+$.
All 37 vector lines questions with mark schemes →
FAQ
How many IB Maths AA HL vector lines questions are there?
There are 37 exam-style vector lines questions in the AA HL question bank (Paper 1: 24 · Paper 2: 13), graded 7 easy, 11 medium, 15 hard, 4 starter. Every question has a full IB-style mark scheme (M, A and R marks).
Is vector lines on Paper 1 or Paper 2?
Both. In the bank, Paper 1: 24 · Paper 2: 13. Paper 1 is non-calculator, so practise the exact-value algebra as well as the GDC methods.
Where can I get the mark schemes?
Open the AA HL Unit 3 practice page: every question has a step-by-step IB-style mark scheme, and you can photograph your working for instant AI marking. The worked examples on this page are free.
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